University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 118: 33


The proof is detailed in the Work step by step below.

Work Step by Step

$$y=mx+b$$ At any point $(x_0,mx_0+b)$, we have $$y'=\lim_{h\to0}\frac{m(x_0+h)+b-(mx_0+b)}{h}$$ $$y'=\lim_{h\to0}\frac{mx_0+mh+b-mx_0-b}{h}=\lim_{h\to0}\frac{mh}{h}$$ $$y'=\lim_{h\to0}m=m$$ So the tangent line at any point $x=x_0$ will have this form: $$y=mx+n$$ To find $n$, we use the coordinates of the given point $(x_0,mx_0+b)$: $$mx_0+n=mx_0+b$$ $$n=b$$ Therefore, the tangent line of the given line is $y=mx+b$, which is exactly the same line. This means the line $y=mx+b$ is its own tangent line at any point $(x_0,mx_0+b)$.
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